3.218 \(\int \frac{\sin ^8(c+d x)}{(a-b \sin ^4(c+d x))^2} \, dx\)

Optimal. Leaf size=320 \[ \frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 b^{3/2} d \left (\sqrt{a}-\sqrt{b}\right )^{3/2}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 b^{3/2} d \left (\sqrt{a}+\sqrt{b}\right )^{3/2}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{\tan ^5(c+d x)}{4 b d \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac{\tan (c+d x)}{4 b d (a-b)}+\frac{x}{b^2} \]

[Out]

x/b^2 - (a^(1/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^2*d) + (
a^(1/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(8*(Sqrt[a] - Sqrt[b])^(3/2)*b^(3/2)*d) - (a^(
1/4)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^2*d) - (a^(1/4)*ArcT
an[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(8*(Sqrt[a] + Sqrt[b])^(3/2)*b^(3/2)*d) - Tan[c + d*x]/(4*
(a - b)*b*d) + Tan[c + d*x]^5/(4*b*d*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4))

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Rubi [A]  time = 0.446449, antiderivative size = 320, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3217, 1313, 1275, 12, 1122, 1166, 205, 1287, 203} \[ \frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 b^{3/2} d \left (\sqrt{a}-\sqrt{b}\right )^{3/2}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}-\sqrt{b}}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 b^{3/2} d \left (\sqrt{a}+\sqrt{b}\right )^{3/2}}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 b^2 d \sqrt{\sqrt{a}+\sqrt{b}}}+\frac{\tan ^5(c+d x)}{4 b d \left ((a-b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a\right )}-\frac{\tan (c+d x)}{4 b d (a-b)}+\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^8/(a - b*Sin[c + d*x]^4)^2,x]

[Out]

x/b^2 - (a^(1/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] - Sqrt[b]]*b^2*d) + (
a^(1/4)*ArcTan[(Sqrt[Sqrt[a] - Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(8*(Sqrt[a] - Sqrt[b])^(3/2)*b^(3/2)*d) - (a^(
1/4)*ArcTan[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(2*Sqrt[Sqrt[a] + Sqrt[b]]*b^2*d) - (a^(1/4)*ArcT
an[(Sqrt[Sqrt[a] + Sqrt[b]]*Tan[c + d*x])/a^(1/4)])/(8*(Sqrt[a] + Sqrt[b])^(3/2)*b^(3/2)*d) - Tan[c + d*x]/(4*
(a - b)*b*d) + Tan[c + d*x]^5/(4*b*d*(a + 2*a*Tan[c + d*x]^2 + (a - b)*Tan[c + d*x]^4))

Rule 3217

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p)/(1 + ff^2
*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 1313

Int[(((f_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> -Dist
[f^4/(c*d^2 - b*d*e + a*e^2), Int[(f*x)^(m - 4)*(a*d + (b*d - a*e)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] + Dist[(
d^2*f^4)/(c*d^2 - b*d*e + a*e^2), Int[((f*x)^(m - 4)*(a + b*x^2 + c*x^4)^(p + 1))/(d + e*x^2), x], x] /; FreeQ
[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 2]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(f*
(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1)*(b*d - 2*a*e - (b*e - 2*c*d)*x^2))/(2*(p + 1)*(b^2 - 4*a*c)), x] - D
ist[f^2/(2*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1)*Simp[(m - 1)*(b*d - 2*a*e) -
(4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[
p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1122

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d^3*(d*x)^(m - 3)*(a + b*
x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 1)), x] - Dist[d^4/(c*(m + 4*p + 1)), Int[(d*x)^(m - 4)*Simp[a*(m - 3) + b
*(m + 2*p - 1)*x^2, x]*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && Gt
Q[m, 3] && NeQ[m + 4*p + 1, 0] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sin ^8(c+d x)}{\left (a-b \sin ^4(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^8}{\left (1+x^2\right ) \left (a+2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+a x^2\right )}{\left (a+2 a x^2+(a-b) x^4\right )^2} \, dx,x,\tan (c+d x)\right )}{b d}-\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right ) \left (a+2 a x^2+(a-b) x^4\right )} \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac{\tan ^5(c+d x)}{4 b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int -\frac{2 a b x^4}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{8 a b^2 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{b \left (1+x^2\right )}+\frac{a \left (1+x^2\right )}{b \left (a+2 a x^2+(a-b) x^4\right )}\right ) \, dx,x,\tan (c+d x)\right )}{b d}\\ &=\frac{\tan ^5(c+d x)}{4 b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{b^2 d}-\frac{a \operatorname{Subst}\left (\int \frac{1+x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{b^2 d}-\frac{\operatorname{Subst}\left (\int \frac{x^4}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{4 b d}\\ &=\frac{x}{b^2}-\frac{\tan (c+d x)}{4 (a-b) b d}+\frac{\tan ^5(c+d x)}{4 b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )}-\frac{\left (a \left (1-\frac{\sqrt{b}}{\sqrt{a}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 b^2 d}-\frac{\left (a \left (1+\frac{\sqrt{b}}{\sqrt{a}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{2 b^2 d}+\frac{\operatorname{Subst}\left (\int \frac{a+2 a x^2}{a+2 a x^2+(a-b) x^4} \, dx,x,\tan (c+d x)\right )}{4 (a-b) b d}\\ &=\frac{x}{b^2}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}-\sqrt{b}} b^2 d}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}+\sqrt{b}} b^2 d}-\frac{\tan (c+d x)}{4 (a-b) b d}+\frac{\tan ^5(c+d x)}{4 b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )}+\frac{\left (\sqrt{a} \left (\sqrt{a}+\sqrt{b}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{8 \left (\sqrt{a}-\sqrt{b}\right ) b^{3/2} d}+\frac{\left (\sqrt{a} \left (2 \sqrt{a}-\frac{a+b}{\sqrt{b}}\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a-\sqrt{a} \sqrt{b}+(a-b) x^2} \, dx,x,\tan (c+d x)\right )}{8 (a-b) b d}\\ &=\frac{x}{b^2}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}-\sqrt{b}} b^2 d}+\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}-\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \left (\sqrt{a}-\sqrt{b}\right )^{3/2} b^{3/2} d}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 \sqrt{\sqrt{a}+\sqrt{b}} b^2 d}-\frac{\sqrt [4]{a} \tan ^{-1}\left (\frac{\sqrt{\sqrt{a}+\sqrt{b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{8 \left (\sqrt{a}+\sqrt{b}\right )^{3/2} b^{3/2} d}-\frac{\tan (c+d x)}{4 (a-b) b d}+\frac{\tan ^5(c+d x)}{4 b d \left (a+2 a \tan ^2(c+d x)+(a-b) \tan ^4(c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 4.95567, size = 262, normalized size = 0.82 \[ \frac{-\frac{\sqrt{a} \left (4 \sqrt{a}+5 \sqrt{b}\right ) \tan ^{-1}\left (\frac{\left (\sqrt{a}+\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}+a}}\right )}{\left (\sqrt{a}+\sqrt{b}\right ) \sqrt{\sqrt{a} \sqrt{b}+a}}+\frac{2 a b (\sin (4 (c+d x))-6 \sin (2 (c+d x)))}{(a-b) (8 a+4 b \cos (2 (c+d x))-b \cos (4 (c+d x))-3 b)}+\frac{\sqrt{a} \left (4 \sqrt{a}-5 \sqrt{b}\right ) \tanh ^{-1}\left (\frac{\left (\sqrt{a}-\sqrt{b}\right ) \tan (c+d x)}{\sqrt{\sqrt{a} \sqrt{b}-a}}\right )}{\left (\sqrt{a}-\sqrt{b}\right ) \sqrt{\sqrt{a} \sqrt{b}-a}}+8 (c+d x)}{8 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^8/(a - b*Sin[c + d*x]^4)^2,x]

[Out]

(8*(c + d*x) - (Sqrt[a]*(4*Sqrt[a] + 5*Sqrt[b])*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + Sqrt[a]*Sqr
t[b]]])/((Sqrt[a] + Sqrt[b])*Sqrt[a + Sqrt[a]*Sqrt[b]]) + (Sqrt[a]*(4*Sqrt[a] - 5*Sqrt[b])*ArcTanh[((Sqrt[a] -
Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b]]])/((Sqrt[a] - Sqrt[b])*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + (2*a*b
*(-6*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/((a - b)*(8*a - 3*b + 4*b*Cos[2*(c + d*x)] - b*Cos[4*(c + d*x)])))/
(8*b^2*d)

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Maple [B]  time = 0.125, size = 644, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^2,x)

[Out]

-1/2/d*a/b/(tan(d*x+c)^4*a-tan(d*x+c)^4*b+2*a*tan(d*x+c)^2+a)/(a-b)*tan(d*x+c)^3-1/4/d*a/b/(tan(d*x+c)^4*a-tan
(d*x+c)^4*b+2*a*tan(d*x+c)^2+a)/(a-b)*tan(d*x+c)-1/2/d*a^2/b^2/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b
)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))+3/4/d*a/b/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x
+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))-3/8/d*a^2/b/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*ta
n(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))+5/8/d*a/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*t
an(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^(1/2))-1/2/d*a^2/b^2/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(
d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+3/4/d*a/b/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(d*x+c)/
(((a*b)^(1/2)-a)*(a-b))^(1/2))+3/8/d*a^2/b/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*tan(
d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))-5/8/d*a/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^(1/2)*arctanh((-a+b)*t
an(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))+1/d/b^2*arctan(tan(d*x+c))

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 9.06827, size = 8038, normalized size = 25.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^2,x, algorithm="fricas")

[Out]

1/32*(32*(a*b - b^2)*d*x*cos(d*x + c)^4 - 64*(a*b - b^2)*d*x*cos(d*x + c)^2 - 32*(a^2 - 2*a*b + b^2)*d*x + ((a
*b^3 - b^4)*d*cos(d*x + c)^4 - 2*(a*b^3 - b^4)*d*cos(d*x + c)^2 - (a^2*b^2 - 2*a*b^3 + b^4)*d)*sqrt(-((a^3*b^4
- 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^
7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) + 16*a^3 - 47*a^2*b + 35*a*b^2
)/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2))*log(32*a^3 - 166*a^2*b + 1125/4*a*b^2 - 625/4*b^3 - 1/4*(128*a^
3 - 664*a^2*b + 1125*a*b^2 - 625*b^3)*cos(d*x + c)^2 + 1/2*(2*(2*a^4*b^5 - 9*a^3*b^6 + 15*a^2*b^7 - 11*a*b^8 +
3*b^9)*d^3*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4
*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))*cos(d*x + c)*sin(d*x + c) - (24*a^3*b^2 - 127*a^2*b^
3 + 220*a*b^4 - 125*b^5)*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2*sqrt((6
4*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10
+ 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) + 16*a^3 - 47*a^2*b + 35*a*b^2)/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*
d^2)) + 1/4*(2*(16*a^4*b^3 - 73*a^3*b^4 + 123*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2*cos(d*x + c)^2 - (16*a^4*b^3 -
73*a^3*b^4 + 123*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2)*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 62
5*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))) - ((a*b^3 -
b^4)*d*cos(d*x + c)^4 - 2*(a*b^3 - b^4)*d*cos(d*x + c)^2 - (a^2*b^2 - 2*a*b^3 + b^4)*d)*sqrt(-((a^3*b^4 - 3*a^
2*b^5 + 3*a*b^6 - b^7)*d^2*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a
^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) + 16*a^3 - 47*a^2*b + 35*a*b^2)/((a^3
*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2))*log(32*a^3 - 166*a^2*b + 1125/4*a*b^2 - 625/4*b^3 - 1/4*(128*a^3 - 664
*a^2*b + 1125*a*b^2 - 625*b^3)*cos(d*x + c)^2 - 1/2*(2*(2*a^4*b^5 - 9*a^3*b^6 + 15*a^2*b^7 - 11*a*b^8 + 3*b^9)
*d^3*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 -
20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))*cos(d*x + c)*sin(d*x + c) - (24*a^3*b^2 - 127*a^2*b^3 + 220
*a*b^4 - 125*b^5)*d*cos(d*x + c)*sin(d*x + c))*sqrt(-((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2*sqrt((64*a^5 -
464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^
2*b^11 - 6*a*b^12 + b^13)*d^4)) + 16*a^3 - 47*a^2*b + 35*a*b^2)/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2)) +
1/4*(2*(16*a^4*b^3 - 73*a^3*b^4 + 123*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2*cos(d*x + c)^2 - (16*a^4*b^3 - 73*a^3*
b^4 + 123*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2)*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4
)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))) + ((a*b^3 - b^4)*d*
cos(d*x + c)^4 - 2*(a*b^3 - b^4)*d*cos(d*x + c)^2 - (a^2*b^2 - 2*a*b^3 + b^4)*d)*sqrt(((a^3*b^4 - 3*a^2*b^5 +
3*a*b^6 - b^7)*d^2*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 +
15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) - 16*a^3 + 47*a^2*b - 35*a*b^2)/((a^3*b^4 - 3
*a^2*b^5 + 3*a*b^6 - b^7)*d^2))*log(-32*a^3 + 166*a^2*b - 1125/4*a*b^2 + 625/4*b^3 + 1/4*(128*a^3 - 664*a^2*b
+ 1125*a*b^2 - 625*b^3)*cos(d*x + c)^2 + 1/2*(2*(2*a^4*b^5 - 9*a^3*b^6 + 15*a^2*b^7 - 11*a*b^8 + 3*b^9)*d^3*sq
rt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*
b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))*cos(d*x + c)*sin(d*x + c) + (24*a^3*b^2 - 127*a^2*b^3 + 220*a*b^4
- 125*b^5)*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2*sqrt((64*a^5 - 464*a^4
*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 -
6*a*b^12 + b^13)*d^4)) - 16*a^3 + 47*a^2*b - 35*a*b^2)/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2)) + 1/4*(2*
(16*a^4*b^3 - 73*a^3*b^4 + 123*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2*cos(d*x + c)^2 - (16*a^4*b^3 - 73*a^3*b^4 + 12
3*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2)*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*
b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))) - ((a*b^3 - b^4)*d*cos(d*x
+ c)^4 - 2*(a*b^3 - b^4)*d*cos(d*x + c)^2 - (a^2*b^2 - 2*a*b^3 + b^4)*d)*sqrt(((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6
- b^7)*d^2*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*
b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4)) - 16*a^3 + 47*a^2*b - 35*a*b^2)/((a^3*b^4 - 3*a^2*b^5
+ 3*a*b^6 - b^7)*d^2))*log(-32*a^3 + 166*a^2*b - 1125/4*a*b^2 + 625/4*b^3 + 1/4*(128*a^3 - 664*a^2*b + 1125*a
*b^2 - 625*b^3)*cos(d*x + c)^2 - 1/2*(2*(2*a^4*b^5 - 9*a^3*b^6 + 15*a^2*b^7 - 11*a*b^8 + 3*b^9)*d^3*sqrt((64*a
^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 1
5*a^2*b^11 - 6*a*b^12 + b^13)*d^4))*cos(d*x + c)*sin(d*x + c) + (24*a^3*b^2 - 127*a^2*b^3 + 220*a*b^4 - 125*b^
5)*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2*sqrt((64*a^5 - 464*a^4*b + 124
1*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^1
2 + b^13)*d^4)) - 16*a^3 + 47*a^2*b - 35*a*b^2)/((a^3*b^4 - 3*a^2*b^5 + 3*a*b^6 - b^7)*d^2)) + 1/4*(2*(16*a^4*
b^3 - 73*a^3*b^4 + 123*a^2*b^5 - 91*a*b^6 + 25*b^7)*d^2*cos(d*x + c)^2 - (16*a^4*b^3 - 73*a^3*b^4 + 123*a^2*b^
5 - 91*a*b^6 + 25*b^7)*d^2)*sqrt((64*a^5 - 464*a^4*b + 1241*a^3*b^2 - 1450*a^2*b^3 + 625*a*b^4)/((a^6*b^7 - 6*
a^5*b^8 + 15*a^4*b^9 - 20*a^3*b^10 + 15*a^2*b^11 - 6*a*b^12 + b^13)*d^4))) - 8*(a*b*cos(d*x + c)^3 - 2*a*b*cos
(d*x + c))*sin(d*x + c))/((a*b^3 - b^4)*d*cos(d*x + c)^4 - 2*(a*b^3 - b^4)*d*cos(d*x + c)^2 - (a^2*b^2 - 2*a*b
^3 + b^4)*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**8/(a-b*sin(d*x+c)**4)**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^8/(a-b*sin(d*x+c)^4)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError